Problem 124 - A Complete Resource Book in Chemistry for JEE Main

Question

A metal oxide has the formula \(\mathrm{Z}_{2} \mathrm{O}_{3}\). It can be reduced by hydrogen to give free metal and water. \(0.1596 \mathrm{~g}\) of the metal oxide requires \(6 \mathrm{mg}\) of hydrogen for complete reduction. The atomic weight of the metal is (a) \(55.8\) (b) \(65.8\) (c) \(6.58\) (d) \(15.9\)

Step-by-Step Solution

Answer

The atomic weight of the metal is (a) 55.8 u.

Step 1: Understand the Reaction

The reduction reaction can be represented as \( \text{Z}_2\text{O}_3 + 3\text{H}_2 \rightarrow 2\text{Z} + 3\text{H}_2\text{O} \). In this process, the hydrogen gas reacts with the metal oxide to yield the metal and water.

Step 2: Calculate Moles of Hydrogen

Given the mass of hydrogen used is \(6 \text{ mg} = 0.006 \text{ g}\). The molar mass of hydrogen (\(\text{H}_2\)) is \(2 \text{ g/mol}\). Therefore, the moles of hydrogen used can be calculated as \( \frac{0.006}{2} = 0.003 \text{ moles} \).

Step 3: Relate Moles of Reactants

From the balanced equation, \(3\) moles of \(\text{H}_2\) are needed to reduce \(1\) mole of \(\text{Z}_2\text{O}_3\). So, \(0.003\) moles of \(\text{H}_2\) will reduce \(\frac{0.003}{3} = 0.001\) moles of \(\text{Z}_2\text{O}_3\).

Step 4: Determine Molar Mass of \(\text{Z}_2\text{O}_3\)

The mass of \(\text{Z}_2\text{O}_3\) given is \(0.1596 \text{ g}\). Using the moles calculated, the molar mass is \( \frac{0.1596}{0.001} = 159.6 \text{ g/mol} \).

Step 5: Calculate Atomic Weight of Metal, \(\text{Z}\)

The molar mass of \(\text{Z}_2\text{O}_3\) is composed of two metal atoms and three oxygen atoms. The molar mass is given by \(2x + 48 = 159.6\), where \(x\) is the atomic weight of \(\text{Z}\), and the mass of three oxygen atoms is \(3 \times 16 = 48\). Solve for \(x\): \(2x = 159.6 - 48 = 111.6\) and \(x = \frac{111.6}{2} = 55.8 \text{ u}\).

Step 6: Verify the Correct Option

The calculated atomic weight of the metal \(\text{Z}\) is \(55.8\). Therefore, the correct answer from the options given is (a) \(55.8\).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It uses the coefficients in a balanced chemical equation to determine the proportions of molecules involved. For the reduction of the metal oxide \(\text{Z}_2\text{O}_3\) by hydrogen, the balanced equation is \(\text{Z}_2\text{O}_3 + 3\text{H}_2 \rightarrow 2\text{Z} + 3\text{H}_2\text{O}\). This equation shows that one mole of the metal oxide reacts with three moles of hydrogen gas to produce two moles of the metal and three moles of water. By understanding stoichiometry, we can predict how much of each substance is needed or produced in a reaction, which is crucial for solving problems that require precise calculations in chemistry.

Molar Mass Calculation

Calculating the molar mass is essential for converting between the mass of a substance and the amount of substance in moles. To find the molar mass of a compound like \(\text{Z}_2\text{O}_3\), we sum the atomic masses of all the atoms in its formula. Use the periodic table to find the atomic weights.

• For oxygen (\(\text{O}\)): 16 u per atom, with three atoms contributing \(3 \times 16 = 48\) u to the total.
• For the metal (\(\text{Z}\)), its atomic weight is initially unknown.

Given that the molar mass for \(\text{Z}_2\text{O}_3\) is calculated to be 159.6 g/mol, we can set up the equation \(2x + 48 = 159.6\), where \(x\) is the atomic weight of \(\text{Z}\). Solving this, we find that \(2x = 111.6\), hence \(x = 55.8\) u, determining the metal's atomic weight.

Atomic Weight Determination

Determining atomic weights is critical for identifying elements and understanding their properties. In the context of the exercise, we find the atomic weight of element \(\text{Z}\) in the compound \(\text{Z}_2\text{O}_3\) by using molar masses and stoichiometry. Once the molar mass of the compound was found, the equation \(2x + 48 = 159.6\) allowed us to isolate \(x\), representing the atomic weight of \(\text{Z}\). Knowing this is vital to solving the problem as the atomic weight directly corresponds to the correct multiple-choice answer. Once solved, it was clear that \(x = 55.8\) u, confirming our calculations and understanding of the compound's makeup.

Hydrogen Reduction Reaction

Hydrogen reduction reactions involve hydrogen gas reducing metal compounds to pure metals. In the exercise, hydrogen gas reacts with the metal oxide \(\text{Z}_2\text{O}_3\) under the equation \(\text{Z}_2\text{O}_3 + 3\text{H}_2 \rightarrow 2\text{Z} + 3\text{H}_2\text{O}\).

• Firstly, it is crucial to remember that hydrogen here serves as a reducing agent, meaning it donates electrons to \(\text{Z}_2\text{O}_3\), breaking its bonds with oxygen and forming water.
• This type of chemical reaction is an example of a redox reaction, specifically focusing on the reduction of the metal compound.
• Analyzing the reduction helps us understand the link between the consumed hydrogen and produced metal, allowing us to quantify reactants and products precisely.

By calculating the moles of \(\text{H}_2\) used and aligning it with stoichiometric principles, we accurately predict the amount of metal produced.